Adanya kenaikan suhu akan memperbesar energi kinetik partikel sehingga memperlemah gaya antar molekul gas terlarut , maka partikel gas tersebut mudah terlepas.

Untuk mengendapkan suatu zat dari larutannya prinsip yang paling mudah dilakukan adalah dengan mereaksikannya dengan kation/anion yang dapat membentuk garam yang sukar larut.

Salah satu faktor yang mempengaruhi kelarutan garam perak phosphat adalah pengaruh ion sejenis. Dengan adanya ion sejenis memperkecil kelarutan garam tersebut.

$\mbox{NaCl}+\mbox{AgNO}_{3}$$\rightarrow\mbox{AgCl}+\mbox{NaNO}_{3}$

$\begin{alignedat}{1}n\mbox{AgNO}_{3} & =\frac{0,17}{170}\mbox{ mol}\\
& =10^{-3}\mbox{mol}\\
{}[\mbox{AgNO}_{3}] & =\frac{10^{-3}}{0,1}\mbox{ M}\\
& =10^{-4}\mbox{ M}\\
\mbox{AgCl} & \rightleftarrows\mbox{Ag}^{+}+\mbox{Cl}^{-}\\
Q_{C} & =[\mbox{Ag}^{+}][\mbox{Cl}^{-}]\\
& =(10^{-4})(0,01)\\
& =10^{-6}
\end{alignedat}
$

Karena nilai $Q_{C}>K_{sp}$, sehingga terjadi endapan garam $\mbox{AgCl}$.

$\begin{array}{ccccc}
\mbox{AgCl} & \rightleftarrows & \mbox{Ag}^{+} & + & \mbox{Cl}^{-}\\
s & & s & & s
\end{array}$

$\begin{alignedat}{1}K_{sp}\mbox{AgCl} & =[\mbox{Ag}^{+}][\mbox{Cl}^{-}]\\
10^{-10} & =(s)(s)\\
10^{-10} & =s^{2}\\
s & =\sqrt{10^{-10}}\\
& =10^{-5}\mbox{M}
\end{alignedat}
$

$\begin{array}{ccccc}
\mbox{AgBr} & \rightleftarrows & \mbox{Ag}^{+} & + & \mbox{Br}^{-}\\
s & & s & & s
\end{array}$

$\begin{alignedat}{1}K_{sp}\mbox{AgBr} & =[\mbox{Ag}^{+}][\mbox{Br}^{-}]\\
5\times10^{-13} & =(s)(s)\\
5\times10^{-13} & =s^{2}\\
s & =\sqrt{5\times10^{-13}}\\
& =7\times10^{-7}\mbox{M}
\end{alignedat}
$

$\begin{array}{ccccc}
\mbox{AgIO}_{3} & \rightleftarrows & \mbox{Ag}^{+} & + & \mbox{IO}_{3}^{-}\\
s & & s & & s
\end{array}$

$\begin{alignedat}{1}K_{sp}\mbox{ AgIO}_{3} & =[\mbox{Ag}^{+}][\mbox{IO}_{3}^{-}]\\
2\times10^{-10} & =(s)(s)\\
2\times10^{-10} & =s^{2}\\
s & =\sqrt{2\times10^{-10}}\\
& =1,4\times10^{-5}\mbox{M}
\end{alignedat}
$

$\begin{array}{ccccc}
\mbox{ Ag}_{2}\mbox{CrO}_{4} & \rightleftarrows & \mbox{2Ag}^{+} & + & \mbox{CrO}_{4}^{2-}\\
s & & 2s & & s
\end{array}$

$\begin{alignedat}{1}K_{sp}\mbox{Ag}_{2}\mbox{CrO}_{4} & =[\mbox{Ag}^{+}]^{2}[\mbox{CrO}_{4}^{2-}]\\
3,2\times10^{-12} & =(2s)^{2}(s)\\
3,2\times10^{-12} & =4s^{3}\\
s & =\sqrt[3]{\frac{3,2\times10^{-12}}{4}}\\
& =9,3\times10^{-5}\mbox{M}
\end{alignedat}
$

$\begin{array}{ccccc}
\mbox{AgI} & \rightleftarrows & \mbox{Ag}^{+} & + & \mbox{I}^{-}\\
s & & s & & s
\end{array}$

$\begin{alignedat}{1}K_{sp}\mbox{ AgI} & =[\mbox{Ag}^{+}][\mbox{I}^{-}]\\
10^{-12} & =(s)(s)\\
10^{-12} & =s^{2}\\
s & =\sqrt{10^{-12}}\\
& =10^{-6}\mbox{M}
\end{alignedat}
$

Kelarutan yang paling besar adalah garam $\mbox{AgCl}$.

1) Kelarutan garam $\mbox{AB}_{2}$ dalam larutan $\mbox{A(OH)}_{2}$ 0,01 M adalah $4\times10^{-4}$

2) Ke dalam larutan $\mbox{AB}_{2}$ dimasukkan larutan $\mbox{ACl}_{2}$ 0,1 M sehingga kelarutan garam $\mbox{AB}_{2}$ semakin kecil

3) Perhitungan $K_{sp}$ menggunakan persamaan $K_{sp}=[\mbox{A}^{2+}][\mbox{B}^{-}]^{2}$

4) Kelarutan $\mbox{AB}_{2}$ dalam air tergantung konsentrasi ion $\mbox{A}^{2+}$ dan $\mbox{B}^{-}$ di dalam larutan.

$\begin{alignedat}{1}[\mbox{MgCl}_{2}] & =\frac{\frac{0,038}{95}}{0,1}\mbox{ M}\\
& =4\times10^{-3}\mbox{M}\\
{}[\mbox{Mg}^{2+}] & =4\times10^{-3}\mbox{M}\\
\mbox{Mg(OH)}_{2} & \rightleftharpoons\mbox{Mg}^{2+}+\mbox{2OH}^{-}\\
K_{sp}\mbox{Mg(OH)}_{2} & =[\mbox{Mg}^{2+}][\mbox{OH}^{-}]^{2}\\
4\times10^{-12} & =4\times10^{-3}[\mbox{OH}^{-}]^{2}\\
{}[\mbox{OH}^{-}] & =10^{-4,5}\\
pOH & =4,5\\
pH & =9,5.
\end{alignedat}
$

$\begin{alignedat}{1}\mbox{mol}\mbox{Ag}_{2}\mbox{S} & =\frac{3,6\times10^{-3}}{248}\mbox{ mol}\\
& =1,2\times10^{-5}\mbox{mol}\\
{}[\mbox{Ag}_{2}\mbox{S}] & =\frac{2\times10^{-6}\mbox{mol}}{0,2\mbox{ L}}\\
& =7,2\times10^{-5}\mbox{M}
\end{alignedat}
$

$\begin{array}{c}
\mbox{Ag}_{2}\mbox{S}\\
7,2\times10^{-5}\mbox{M}
\end{array}$$\begin{array}{cc}
\rightleftharpoons & \mbox{2Ag}^{+}\\
& 1,44\times10^{-4}\mbox{M}
\end{array}$$\begin{array}{cc}
+ & \mbox{S}^{2-}\\
& 7,2\times10^{-5}\mbox{M}
\end{array}$

$\begin{alignedat}{1}K_{sp} & =[\mbox{Ag}^{+}]^{2}[\mbox{S}^{2-}]\\
& =(1,44\times10^{-4})^{2}(7,2\times10^{-5})\\
& =1,5\times10^{-13}.
\end{alignedat}
$

$\begin{array}{ccccc}
\mbox{Bi}_{2}\mbox{S}_{3} & \rightleftharpoons & \mbox{2Bi}^{3+} & + & \mbox{3S}^{2-}\\
s & & 2s & & 3s\\
\mbox{Bi}\left(\mbox{NO}_{3}\right)_{3} & \rightarrow & \mbox{Bi}^{3+} & + & 3\mbox{NO}_{3}^{-}\\
0,2\mbox{ M} & & 0,2\mbox{ M} & & 0,4\mbox{ M}
\end{array}$

$\begin{aligned}K_{sp}\mbox{Bi}_{2}\mbox{S}_{3} & =[\mbox{Bi}^{3+}]^{2}[\mbox{S}^{2-}]^{3}\\
1,6\times10^{-72} & =(0,2)^{2}(3s)^{3}\\
1,6\times10^{-72} & =1,08s^{5}\\
s & =\sqrt[5]{\frac{1,6\times10^{-72}}{1,08}}\\
& =4,3\times10^{-15}\mbox{M}.
\end{aligned}
$

$\begin{alignedat}{1}\mbox{mol AgNO}_{3} & =0,3\times0,1\mbox{ mol}\\
& =0,03\mbox{ mol}\\
\mbox{mol}\mbox{Ca}\left(\mbox{NO}_{3}\right)_{2} & =\frac{1,11}{111}\mbox{ mol}\\
& =0,01\mbox{ mol}
\end{alignedat}
$

$\begin{array}{cc}
& \mbox{2AgNO}_{3}\\
\mbox{m} & 0,03\mbox{ mol}\\
\mbox{r} & 0,02\mbox{ mol}\\
\mbox{s} & 0,01\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & \mbox{CaCl}_{2}\\
& 0,01\mbox{ mol}\\
& 0,01\mbox{ mol}\\
& –
\end{array}$$\begin{array}{cc}
\rightarrow & \mbox{2AgCl}\\
\\
& 0,02\mbox{ mol}\\
& 0,02\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & \mbox{Ca}\left(\mbox{NO}_{3}\right)_{2}\\
\\
& 0,01\mbox{ mol}\\
& 0,01\mbox{ mol}
\end{array}$

Jumlah konsentrasi ion $\mbox{NO}_{3}^{-}$ adalah:

$\begin{array}{c}
\mbox{AgNO}_{3}\\
0,01\mbox{ mol}\\
\mbox{Ca}\left(\mbox{NO}_{3}\right)_{2}\\
0,01\mbox{ mol}
\end{array}$$\begin{array}{cc}
\rightarrow & \mbox{Ag}^{+}\\
& 0,01\mbox{ mol}\\
\rightarrow & \mbox{Ca}^{2+}\\
& 0,01\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & \mbox{NO}_{3}^{-}\\
& 0,01\mbox{ mol}\\
+ & \mbox{2NO}_{3}^{-}\\
& 0,02\mbox{ mol}
\end{array}$

$\begin{alignedat}{1}[\mbox{NO}_{3}^{-}]_{\mbox{total}} & =\frac{0,01+0,02}{0,1}\mbox{M}\\
& =0,3\mbox{ M}.
\end{alignedat}
$

Yang dapat bereaksi dengan gas karbon dioksida hanyalah $\mbox{X(OH)}_{2}$ pada larutannya. Maka kita memerlukan konsentrasi yang lebih kecil dari:

$\begin{aligned}K_{sp} & =4s^{3}\\
s & =0,05\mbox{ M}
\end{aligned}
$

Pada konsentrasi ini larutan jenuh, sehingga kita pilih konsentrasi yang lebih rendah dari titik jenuh ini dimana $\mbox{X(OH)}_{2}$ masih larut dan dapat bereaksi dengan karbondioksida.

$\mbox{Ag}_{2}\mbox{CrO}_{4}\rightleftharpoons\mbox{2Ag}^{+}+\mbox{CrO}_{4}^{2-}$

$\begin{array}{cccc}
K_{sp}\mbox{Ag}_{2}\mbox{CrO}_{4} & = & [\mbox{Ag}^{+}]^{2} & [\mbox{CrO}_{4}^{2-}]\\
3,2\times10^{-12} & = & (0,01)^{2} & [\mbox{CrO}_{4}^{2-}]\\
{}[\mbox{CrO}_{4}^{2-}] & = & 3,2\times10^{-8}\mbox{M}
\end{array}$

$\mbox{BaCrO}_{4}\rightleftharpoons\mbox{Ba}^{2+}+\mbox{CrO}_{4}^{2-}$

$\begin{array}{cccc}
K_{sp}\mbox{BaCrO}_{4} & = & [\mbox{Ba}^{2+}] & [\mbox{CrO}_{4}^{2-}]\\
1\times10^{-10} & = & (0,01)^{2} & [\mbox{CrO}_{4}^{2-}]\\
{}[\mbox{CrO}_{4}^{2-}] & = & 1\times10^{-6}\mbox{M}
\end{array}$

Untuk mengendapkan $\mbox{Ag}_{2}\mbox{CrO}_{4}$ diperlukan $[\mbox{CrO}_{4}^{2-}]$ sejumlah $3,2\times10^{-8}\mbox{M}$ sedangkan untuk mengendapkan $\mbox{BaCrO}_{4}$ diperlukan $[\mbox{CrO}_{4}^{2-}]$ sejumlah $1\times10^{-6}\mbox{M}$. Sehingga garam yang mengendap terlebih dahulu adalah $\mbox{Ag}_{2}\mbox{CrO}_{4}$.
Sehingga konsentrasi $[\mbox{Ba}^{2+}]$$=0,01$ M.

$\mbox{Mg(OH)}_{2}$$\rightleftharpoons\mbox{Mg}^{2+}+\mbox{2OH}^{-}$

$\begin{array}{c}
\mbox{Mg}\left(\mbox{NO}_{3}\right)_{2}\\
0,003\mbox{ mol}
\end{array}$$\begin{array}{cc}
\rightarrow & \mbox{Mg}^{2+}\\
& 0,003\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & \mbox{2NO}_{3}^{-}\\
& 0,006\mbox{ mol}
\end{array}$

$\begin{alignedat}{1}K_{sp} & =[\mbox{Mg}^{2+}][\mbox{OH}^{-}]^{2}\\
1,12\times10^{-11} & =(\frac{0,003}{2})[\mbox{OH}^{-}]^{2}\\
{}[\mbox{OH}^{-}]^{2} & =7,5\times10^{-9}\mbox{M}^{2}\\
{}[\mbox{OH}^{-}] & =8,6\times10^{-5}\mbox{M}
\end{alignedat}
$

Campuran yang terdiri atas amonia dan amonium klorida merupakan larutan buffer, sehingga

$\begin{alignedat}{1}[\mbox{OH}^{-}] & =K_{b}\times\frac{\mbox{mol basa}}{\mbox{mol asam konjugat}}\\
8.6\times10^{-5}\mbox{M} & =1,8\times10^{-5}\times\frac{0,02\times2}{\mbox{mol NH}_{4}\mbox{Cl}}\\
\mbox{mol NH}_{4}\mbox{Cl} & =119,4\mbox{ mol}.
\end{alignedat}
$

$\begin{array}{cc}
& \mbox{2AgNO}_{3}\\
& 0,3\times0,01\mbox{ mol}\\
\mbox{m} & 0,003\mbox{ mol}\\
\mbox{r} & 0,002\mbox{ mol}\\
\mbox{s} & 0,001\mbox{ mol}
\end{array}$ $\begin{array}{cc}
+ & \mbox{Na}_{2}\mbox{CrO}_{4}\\
& 0,2\times0,05\mbox{ mol}\\
& 0,001\mbox{ mol}\\
& 0,001\mbox{ mol}\\
& –
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{Ag}_{2}\mbox{CrO}_{4}\\
\\
\\
& 0,001\mbox{ mol}\\
& 0,001\mbox{ mol}
\end{array}$ $\begin{array}{cc}
+ & \mbox{2NaNO}_{3}\\
\\
\\
& 0,002\mbox{ mol}\\
& 0,002\mbox{ mol}
\end{array}$

Menentukan kelarutan ion $\mbox{Ag}^{+}$ dari garam yang sukar larut $\mbox{Ag}_{2}\mbox{CrO}_{4}$ dengan adanya ion sejenis dari sisa reaktan $\mbox{AgNO}_{3}$.

$\begin{array}{c}
\mbox{AgNO}_{3}\\
0,001\mbox{ mol}
\end{array}$$\begin{array}{cc}
\rightarrow & \mbox{Ag}^{+}\\
& 0,001\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & \mbox{NO}_{3}^{-}\\
& 0,001\mbox{ mol}
\end{array}$

$\begin{alignedat}{1}\mbox{Ag}_{2}\mbox{CrO}_{4} & \rightleftharpoons\mbox{2Ag}^{+}+\mbox{CrO}_{4}^{2-}\\
K_{sp}\mbox{Ag}_{2}\mbox{CrO}_{4} & =[\mbox{Ag}^{+}]^{2}[\mbox{CrO}_{4}^{2-}]\\
3,2\times10^{-12} & =(\frac{0,001}{(0,3+0,2)})[\mbox{CrO}_{4}^{2-}]\\
{}[\mbox{CrO}_{4}^{2-}] & =1,6\times10^{-9}\mbox{M}
\end{alignedat}
$

$\begin{array}{c}
\mbox{Ag}_{2}\mbox{CrO}_{4}\\
1,6\times10^{-9}\mbox{M}
\end{array}$$\begin{array}{cc}
\rightleftharpoons & \mbox{2Ag}^{+}\\
& 3,2\times10^{-9}\mbox{M}
\end{array}$$\begin{array}{cc}
+ & \mbox{CrO}_{4}^{2-}\\
& 1,6\times10^{-9}\mbox{M}
\end{array}$

$\begin{alignedat}{1}\mbox{massa}\mbox{Ag}^{+} & =M\times V\times A_{r}\mbox{ Ag}\\
& =2\times10^{-3}\times0,5\times108\mbox{ gram}\\
& =0,108\mbox{ gram.}
\end{alignedat}
$

$\begin{array}{cc}
& \mbox{Ba(OH)}_{2}\\
& 200\mbox{ mL}\times0,01\mbox{ M}\\
\mbox{m} & 2\mbox{ mmol}\\
\mbox{r} & 2\mbox{ mmol}\\
\mbox{s} & –
\end{array}$$\begin{array}{cc}
+ & 2\mbox{CH}_{3}\mbox{COOH}\\
& 200\mbox{ mL}\times0,02\mbox{ M}\\
& 4\mbox{ mmol}\\
& 4\mbox{ mmol}\\
& –
\end{array}$$\begin{array}{cc}
\rightarrow & \left(\mbox{CH}_{3}\mbox{COO}\right)_{2}\mbox{Ba}\\
\\
\\
& 2\mbox{ mmol}\\
& 2\mbox{ mmol}
\end{array}$$\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\\
& 4\mbox{ mmol}\\
& 4\mbox{ mmol}
\end{array}$

$\begin{alignedat}{1}\mbox{Fe(OH)}_{2} & \rightleftharpoons\mbox{Fe}^{2+}+\mbox{2OH}^{-}\\
K_{sp}\mbox{Fe(OH)}_{2} & =[\mbox{Fe}^{2+}][\mbox{OH}^{-}]^{2}\\
6\times10^{-13} & =(0,06)[\mbox{OH}^{-}]^{2}\\
{}[\mbox{OH}^{-}] & =10^{-5,5}\mbox{M}\\
{}[\mbox{OH}^{-}] & =\sqrt{\frac{K_{W}\times[G]\times val}{K_{a}}}\\
10^{-5,5} & =\sqrt{\frac{10^{-14}\times\frac{2\mbox{ mmol}}{400\mbox{ mL}}\times2}{K_{a}}}\\
10^{-11} & =\frac{10^{-14}\times\frac{2\mbox{ mmol}}{400\mbox{ mL}}\times2}{K_{a}}\\
K_{a} & =10^{-5}.
\end{alignedat}
$